Thursday, January 31, 2013

1. The positive integers are colored black and white. The sum of two differently colorednumbers is black, and their product is white. What is the product of two whitenumbers? Find all such colorings.

2. Each element of a 25 × 25 matrix is either +1 or −1. Let ai be the product of allelements of the i-th row and bj be the product of all elements of the j-th column.Prove that a1 + b1 + · · · + a25 + b25
≠ 0.

Sunday, January 27, 2013

Problem: Prove that an a × b rectangle can be covered by 1 × n rectangles iff n|a or n|b.

Solution: The proof is given by means of the following 2 claims.

Claim 1:
Given a
rectangular area is covered by tiles of dimension 1xn placed in horizontal or
vertical orientation,
it is
possible to remove the
tiles
one at a time in such a way that at any point of time, the covered region’s top
border consists of a sequence of steps increasing from left to right as shown
in diagram 1.

Proof:
Define step
blocks to
be rectangular regions defined by the step-like shape of the covered area. In
diagram 1, step
blocks are
represented by the areas shaded blue. Start from the left most step block and
consider the top left tile in that block. Thistile must fall entirely within the current
block (call it block A) or must extend beyond the
right of the current
step block
(it cannot beyond the bottom of the current block since the first block is also
the bottommost). In the former
case, removing the
tile
will leave behind a step shape,
thereby proving the claim. In the latter case, inspect the next step block for
the possibility of removing a tile. The top left tile of this next block (call
it block B) cannot extend beyond the bottom of that block since the block A’s
top left tile occupies that portion as per the previous statement. Hence the
top left tile of block B too must either lie entirely within B or extend beyond
the right of B. As before, in the former case, removing the tile will leave a
step shape whereas in the latter case, the search for a tile to remove can be
repeated with the block to the right of C. Note that this procedure can be
repeated at most till the rightmost block is
reached, in which case the top left tile cannot extend beyond the right of the
block (and must therefore lie entirely within that block), thereby becoming
eligible for removal while retaining the step shape.

Claim 2: When
tiles are removed as per the procedure of claim 1 and neither a not b are
divisible by n, then at any point of time at least one of the step regions on
the top border will have a height and a width (as defined by diagram 2) that are
both indivisible by n.

Proof: The
proof is by induction on r, the number of tiles removed so far under the
procedure of claim 1. When r is 0, the covered region and also the only step
region is the original axb
rectangle. The height and width, namely a and b, are both indivisible by n as
per the assumption. Suppose the claim is true for values of r upto R.
Now consider r = (R + 1). Suppose the tile being removed at the current
iteration lies in a step region whose width is W and height Is H. If either W
or H is divisible by n, then there must be some other step region with width
and height indivisible by n and that region will remain after iteration (R +
1)as well. But if say W and H are both
indivisible by n. After removing the tile in this iteration, the 2 new step
regions would have dimensions (a) (W, H - 1) and (W - n, H) or (b) (W, H
- n) and (W - 1, H), depending on the orientation of the tile; in either case,
one of the new step regions would have its width and height indivisible by n.
Hence proved.

Diagram 1:Step
blocks denoted
by shaded blue regions

Diagram 2:
Width of step regions are denoted by L1 and
L2; heights are denoted by H1 and H2. The
rectanglebounded by blue and red lines
is the top left tile of its step block and is eligible for removal as
per the procedure of claim 1

Saturday, January 26, 2013

Divisibility puzzle: Suppose that there are n integers which have the followingproperty: the difference between the product of any n - 1 integers andthe remaining one is divisible by n . Prove that the sum of the square ofthese n numbers is also divisible by n.

Solution:

Denote the n integers by a1, a2, a3, .., an. By the given property, it follows that:

(a1 – a2 a3…an) =
k1n

(a2 – a1 a3 ...an)
= k2n

…

(an – a1 a2… an-1)
= knn

where k1, k2, .., kn are some integers. By multiplying equation 1 above by a1, similarly equation 2 by a2 and so on till the last equation above (which would be multiplied by an), the following set of equations are obtained:

Friday, January 18, 2013

Problem: Every space point is colored with exactly one of the colors red, green or blue. The sets R, G, B consist of the lengths of those segments in space with both end points red, green and blue, respectively. Show that at least one of these sets contains all non-negative real numbers.

Solution: (With reference to the diagram) Suppose the 3 colours are C1, C2 and C3. Assume that the minimum distances by which no two points with the same colour are separated (i.e. no line segment with end points of the same colour exist) be x, y and z for C1, C2 and C3 respectively. Suppose x >= y >= z. Then as shown in the diagram, any point coloured C1 would form the centre of a sphere of radius x such that the sphere can have only points coloured C2 or C3. As indicated in the diagram, a circle of diameter x√3 would then exist on the surface of the sphere such that all its points are coloured C3 and the chords of the circle form line segments of all possible lengths upto x√3, contradicting the assumption that no two points coloured C3 exist at a distance of z from each other.

Diagram for problem #24: The green circle is formed by points on surface of the sphere that are at a distance of x from the top point (on spherical surface) coloured C2, where x is the radius of the sphere centered at a point coloured C1. The diameter of the circle is xv3. If neither of the colours C1 and C2 have 2 points at a distance of x, then the circle must be coloured C3 entirely.